Customer churn in Banking¶

The rate at which a business loses their customers, its more expensive to retain you customers than acquire someone new, there are various ways in which a customer could get churned depending on the nature of the product or service and how the customer is linked to the business.

A customer lifecycle is highlighted below :¶

Full-B2B-customer-lifecycle.webp

The Churn map gives a clear illustration of the diffrent churn pools and the thought process of a customer landing on one of the segments.

Screenshot 2023-10-20 at 1.57.34 AM.png

For this project we will only be predicting whether the customer is a churner or non- churner as a bank customer using a credit card¶

The way we would predict whether a new customer could churn within the next six months boils down to understanding the characteristics of the customers who have churned and modeling that information.

Data loading and preparation¶

The following will be our approach of being data centric until we put out a well furnished model.

Screenshot 2023-10-20 at 2.03.12 AM.png

In [12]:
# Import the packages 
import pandas as pd
import numpy as np
import seaborn as sns
import matplotlib.pyplot as plt
In [13]:
df = pd.read_csv("/Users/maneth/Desktop/Churn_Modelling.csv")
In [14]:
df
Out[14]:
RowNumber CustomerId Surname CreditScore Geography Gender Age Tenure Balance NumOfProducts HasCrCard IsActiveMember EstimatedSalary Exited
0 1 15634602 Hargrave 619 France Female 42 2 0.00 1 1 1 101348.88 1
1 2 15647311 Hill 608 Spain Female 41 1 83807.86 1 0 1 112542.58 0
2 3 15619304 Onio 502 France Female 42 8 159660.80 3 1 0 113931.57 1
3 4 15701354 Boni 699 France Female 39 1 0.00 2 0 0 93826.63 0
4 5 15737888 Mitchell 850 Spain Female 43 2 125510.82 1 1 1 79084.10 0
... ... ... ... ... ... ... ... ... ... ... ... ... ... ...
9995 9996 15606229 Obijiaku 771 France Male 39 5 0.00 2 1 0 96270.64 0
9996 9997 15569892 Johnstone 516 France Male 35 10 57369.61 1 1 1 101699.77 0
9997 9998 15584532 Liu 709 France Female 36 7 0.00 1 0 1 42085.58 1
9998 9999 15682355 Sabbatini 772 Germany Male 42 3 75075.31 2 1 0 92888.52 1
9999 10000 15628319 Walker 792 France Female 28 4 130142.79 1 1 0 38190.78 0

10000 rows × 14 columns

In [15]:
df.head()
Out[15]:
RowNumber CustomerId Surname CreditScore Geography Gender Age Tenure Balance NumOfProducts HasCrCard IsActiveMember EstimatedSalary Exited
0 1 15634602 Hargrave 619 France Female 42 2 0.00 1 1 1 101348.88 1
1 2 15647311 Hill 608 Spain Female 41 1 83807.86 1 0 1 112542.58 0
2 3 15619304 Onio 502 France Female 42 8 159660.80 3 1 0 113931.57 1
3 4 15701354 Boni 699 France Female 39 1 0.00 2 0 0 93826.63 0
4 5 15737888 Mitchell 850 Spain Female 43 2 125510.82 1 1 1 79084.10 0
In [16]:
df.describe()
Out[16]:
RowNumber CustomerId CreditScore Age Tenure Balance NumOfProducts HasCrCard IsActiveMember EstimatedSalary Exited
count 10000.00000 1.000000e+04 10000.000000 10000.000000 10000.000000 10000.000000 10000.000000 10000.00000 10000.000000 10000.000000 10000.000000
mean 5000.50000 1.569094e+07 650.528800 38.921800 5.012800 76485.889288 1.530200 0.70550 0.515100 100090.239881 0.203700
std 2886.89568 7.193619e+04 96.653299 10.487806 2.892174 62397.405202 0.581654 0.45584 0.499797 57510.492818 0.402769
min 1.00000 1.556570e+07 350.000000 18.000000 0.000000 0.000000 1.000000 0.00000 0.000000 11.580000 0.000000
25% 2500.75000 1.562853e+07 584.000000 32.000000 3.000000 0.000000 1.000000 0.00000 0.000000 51002.110000 0.000000
50% 5000.50000 1.569074e+07 652.000000 37.000000 5.000000 97198.540000 1.000000 1.00000 1.000000 100193.915000 0.000000
75% 7500.25000 1.575323e+07 718.000000 44.000000 7.000000 127644.240000 2.000000 1.00000 1.000000 149388.247500 0.000000
max 10000.00000 1.581569e+07 850.000000 92.000000 10.000000 250898.090000 4.000000 1.00000 1.000000 199992.480000 1.000000
In [17]:
df.columns.to_list()
Out[17]:
['RowNumber',
 'CustomerId',
 'Surname',
 'CreditScore',
 'Geography',
 'Gender',
 'Age',
 'Tenure',
 'Balance',
 'NumOfProducts',
 'HasCrCard',
 'IsActiveMember',
 'EstimatedSalary',
 'Exited']
In [8]:
print("Number of Rows:", df.shape[0])
print("Number of Columns:", df.shape[1])

Number of Rows: 10000
Number of Columns: 14
In [10]:
df.info()

<class 'pandas.core.frame.DataFrame'>
RangeIndex: 10000 entries, 0 to 9999
Data columns (total 14 columns):
 #   Column           Non-Null Count  Dtype  
---  ------           --------------  -----  
 0   RowNumber        10000 non-null  int64  
 1   CustomerId       10000 non-null  int64  
 2   Surname          10000 non-null  object 
 3   CreditScore      10000 non-null  int64  
 4   Geography        10000 non-null  object 
 5   Gender           10000 non-null  object 
 6   Age              10000 non-null  int64  
 7   Tenure           10000 non-null  int64  
 8   Balance          10000 non-null  float64
 9   NumOfProducts    10000 non-null  int64  
 10  HasCrCard        10000 non-null  int64  
 11  IsActiveMember   10000 non-null  int64  
 12  EstimatedSalary  10000 non-null  float64
 13  Exited           10000 non-null  int64  
dtypes: float64(2), int64(9), object(3)
memory usage: 1.1+ MB
In [11]:
df.isnull().sum()
Out[11]:
RowNumber          0
CustomerId         0
Surname            0
CreditScore        0
Geography          0
Gender             0
Age                0
Tenure             0
Balance            0
NumOfProducts      0
HasCrCard          0
IsActiveMember     0
EstimatedSalary    0
Exited             0
dtype: int64
In [18]:
df = df.drop(['RowNumber','CustomerId','Surname'],axis=1)
In [19]:
df
Out[19]:
CreditScore Geography Gender Age Tenure Balance NumOfProducts HasCrCard IsActiveMember EstimatedSalary Exited
0 619 France Female 42 2 0.00 1 1 1 101348.88 1
1 608 Spain Female 41 1 83807.86 1 0 1 112542.58 0
2 502 France Female 42 8 159660.80 3 1 0 113931.57 1
3 699 France Female 39 1 0.00 2 0 0 93826.63 0
4 850 Spain Female 43 2 125510.82 1 1 1 79084.10 0
... ... ... ... ... ... ... ... ... ... ... ...
9995 771 France Male 39 5 0.00 2 1 0 96270.64 0
9996 516 France Male 35 10 57369.61 1 1 1 101699.77 0
9997 709 France Female 36 7 0.00 1 0 1 42085.58 1
9998 772 Germany Male 42 3 75075.31 2 1 0 92888.52 1
9999 792 France Female 28 4 130142.79 1 1 0 38190.78 0

10000 rows × 11 columns

In [20]:
df['Geography'].unique()
Out[20]:
array(['France', 'Spain', 'Germany'], dtype=object)

An illustration of the customers retainment per country¶

In [24]:
# Group the data by 'Geography' and 'Exited' and count the number of occurrences
grouped_data = df.groupby(['Geography', 'Exited']).size().unstack(fill_value=0)

# Plot the data
ax = grouped_data.plot(kind='bar', stacked=True)
plt.xlabel('Country')
plt.ylabel('Number of Customers')
plt.title('Number of Customers by Country and Exit Status')
ax.set_xticklabels(ax.get_xticklabels(), rotation=0)

plt.show()
No description has been provided for this image

Analysis¶

France has the highest number of customers churned and the least in Germany, Spain shows more loyalty by the ratio to churned and unchurned customers being low.

Screenshot 2023-10-20 at 2.02.10 AM.png

Encdoing categorical data¶

Preparing data to be fed into the ML model

In [25]:
df = pd.get_dummies(df,drop_first=True)
In [26]:
df
Out[26]:
CreditScore Age Tenure Balance NumOfProducts HasCrCard IsActiveMember EstimatedSalary Exited Geography_Germany Geography_Spain Gender_Male
0 619 42 2 0.00 1 1 1 101348.88 1 False False False
1 608 41 1 83807.86 1 0 1 112542.58 0 False True False
2 502 42 8 159660.80 3 1 0 113931.57 1 False False False
3 699 39 1 0.00 2 0 0 93826.63 0 False False False
4 850 43 2 125510.82 1 1 1 79084.10 0 False True False
... ... ... ... ... ... ... ... ... ... ... ... ...
9995 771 39 5 0.00 2 1 0 96270.64 0 False False True
9996 516 35 10 57369.61 1 1 1 101699.77 0 False False True
9997 709 36 7 0.00 1 0 1 42085.58 1 False False False
9998 772 42 3 75075.31 2 1 0 92888.52 1 True False True
9999 792 28 4 130142.79 1 1 0 38190.78 0 False False False

10000 rows × 12 columns

In [27]:
df.head()
Out[27]:
CreditScore Age Tenure Balance NumOfProducts HasCrCard IsActiveMember EstimatedSalary Exited Geography_Germany Geography_Spain Gender_Male
0 619 42 2 0.00 1 1 1 101348.88 1 False False False
1 608 41 1 83807.86 1 0 1 112542.58 0 False True False
2 502 42 8 159660.80 3 1 0 113931.57 1 False False False
3 699 39 1 0.00 2 0 0 93826.63 0 False False False
4 850 43 2 125510.82 1 1 1 79084.10 0 False True False

Working with the target values¶

In [28]:
df['Exited'].value_counts()
Out[28]:
Exited
0    7963
1    2037
Name: count, dtype: int64
In [37]:
exit_counts = df['Exited'].value_counts()

# Create a bar graph
exit_counts.plot(kind='bar', color=['blue', 'green'])
plt.xlabel('Exit Status (0 = Not Exited, 1 = Exited)')
plt.ylabel('Number of Customers')
plt.title('Total Number of Exited and Not Exited Customers')
plt.xticks([0, 1], ['Not Exited', 'Exited'])
plt.xticks(rotation=45)
plt.show()
No description has been provided for this image

Collecting dependent and independent variables¶

In [38]:
x = df.drop('Exited',axis=1)
x
Out[38]:
CreditScore Age Tenure Balance NumOfProducts ... IsActiveMember EstimatedSalary Geography_Germany Geography_Spain Gender_Male
0 619 42 2 0.00 1 ... 1 101348.88 False False False
1 608 41 1 83807.86 1 ... 1 112542.58 False True False
2 502 42 8 159660.80 3 ... 0 113931.57 False False False
3 699 39 1 0.00 2 ... 0 93826.63 False False False
4 850 43 2 125510.82 1 ... 1 79084.10 False True False
... ... ... ... ... ... ... ... ... ... ... ...
9995 771 39 5 0.00 2 ... 0 96270.64 False False True
9996 516 35 10 57369.61 1 ... 1 101699.77 False False True
9997 709 36 7 0.00 1 ... 1 42085.58 False False False
9998 772 42 3 75075.31 2 ... 0 92888.52 True False True
9999 792 28 4 130142.79 1 ... 0 38190.78 False False False

10000 rows × 11 columns

In [39]:
y = df['Exited']
In [40]:
y
Out[40]:
0       1
1       0
2       1
3       0
4       0
       ..
9995    0
9996    0
9997    1
9998    1
9999    0
Name: Exited, Length: 10000, dtype: int64

Handling Imbalance data using SMOTE¶

In [67]:
!pip3 install -U imbalanced-learn --quiet

101173.07s - pydevd: Sending message related to process being replaced timed-out after 5 seconds

[notice] A new release of pip is available: 23.2.1 -> 23.3
[notice] To update, run: python3.11 -m pip install --upgrade pip
In [115]:
from imblearn.over_sampling import SMOTE
In [116]:
x_res, y_res = SMOTE().fit_resample(x,y)
In [117]:
y_res.value_counts()
Out[117]:
Exited
1    7963
0    7963
Name: count, dtype: int64

Customer-Churn.png

Splitting the data to Training and Testing sets¶

In [118]:
from sklearn.model_selection import train_test_split
In [149]:
x_train,x_test,y_train,y_test = train_test_split(x_res,y_res,test_size=0.2,random_state=42)

Feature Scaling¶

Allows us to put features to the same scale, feature scaling is essential if we are measuring distances with data and not let high values dominate

In [150]:
from sklearn.preprocessing import StandardScaler
In [151]:
sc = StandardScaler()
In [152]:
X_train = sc.fit_transform(x_train)
X_test = sc.fit_transform(x_test)
In [153]:
X_train
Out[153]:
array([[ 0.04702001,  0.60571986,  0.81085019, ...,  1.30571177,
        -0.6342803 ,  0.81396207],
       [ 1.0994971 ,  0.10324089, -1.40111185, ..., -0.76586581,
        -0.6342803 ,  0.81396207],
       [ 1.85596501, -1.00221285, -0.29513083, ...,  1.30571177,
        -0.6342803 ,  0.81396207],
       ...,
       [ 0.20050625,  0.20373669,  1.17951052, ..., -0.76586581,
        -0.6342803 ,  0.81396207],
       [ 0.05798331, -0.39923808,  0.07352951, ..., -0.76586581,
         1.57659003, -1.22855847],
       [ 0.52940534,  1.00770304, -1.40111185, ...,  1.30571177,
        -0.6342803 ,  0.81396207]])

Choosing the ML model¶

Since the target value is of two types, a classification linear model would be a perfect fit and the best out of all the options would be the following models : 1.Logistic Regression model, 2.SVC - Support Vector Classfier, 3.K-Neighbors Classifier

machine-learning-models4.png

Logistic Regression¶

In [154]:
from sklearn.linear_model import LogisticRegression
In [155]:
log = LogisticRegression()
In [156]:
log.fit(X_train, y_train)
Out[156]:
LogisticRegression()
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LogisticRegression()
In [157]:
y_predict = log.predict(X_test)
In [145]:
from sklearn.metrics import accuracy_score
In [146]:
accuracy_score(y_test,y_predict)
Out[146]:
0.809
In [129]:
accuracy_score(y_test,y_predict) # After handling the imbalanced data
Out[129]:
0.7790332705586943
In [148]:
# Since dataset is imbalanced we run the following errors 
from sklearn.metrics import precision_score,recall_score,f1_score
precision = precision_score(y_test, y_predict)
recall = recall_score(y_test, y_predict)
f1_score = f1_score(y_test, y_predict)

print(f"The precision score is: {precision:.3f}") # To correctly predict the postives out of all the postive preds made 
print(f"The recall score is: {recall:.3f}")
print(f"The f1_score score is: {f1_score:.3f}")

The precision score is: 0.594
The recall score is: 0.194
The f1_score score is: 0.293
In [159]:
from sklearn.metrics import precision_score,recall_score,f1_score
precision = precision_score(y_test, y_predict)
recall = recall_score(y_test, y_predict)
f1_score = f1_score(y_test, y_predict)

print("After the imbalanced values were handled with SMOTE")
print(f"The precision score is: {precision:.3f}") # To correctly predict the postives out of all the postive preds made 
print(f"The recall score is: {recall:.3f}")
print(f"The f1_score score is: {f1_score:.3f}")

After the imbalanced values were handled with SMOTE
The precision score is: 0.765
The recall score is: 0.788
The f1_score score is: 0.777

SVC - Support Vector Classfier¶

In [160]:
from sklearn import svm
In [161]:
svm = svm.SVC()
In [171]:
svm.fit(X_train, y_train)
Out[171]:
SVC()
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SVC()
In [172]:
y_predSVC = svm.predict(X_test)
In [173]:
accuracy_score(y_test,y_predSVC)
Out[173]:
0.8465160075329566
In [207]:
precisionSVC = precision_score(y_test, y_predSVC)
print(f"The precision score is: {precisionSVC:.3f}")

The precision score is: 0.843

KNeighbors Classifier¶

In [167]:
from sklearn.neighbors import KNeighborsClassifier
In [168]:
knn = KNeighborsClassifier()
In [169]:
knn.fit(X_train, y_train)
Out[169]:
KNeighborsClassifier()
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KNeighborsClassifier()
In [175]:
y_predKNN = knn.predict(X_test)
In [176]:
accuracy_score(y_test,y_predKNN)
Out[176]:
0.8223477715003139
In [208]:
precisionKNN = precision_score(y_test, y_predKNN)
print(f"The precision score is: {precisionKNN:.3f}")

The precision score is: 0.806

Decision Tree Classifier¶

In [178]:
from sklearn.tree import DecisionTreeClassifier
In [187]:
dt = DecisionTreeClassifier()
In [188]:
dt.fit(X_train,y_train)
Out[188]:
DecisionTreeClassifier()
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DecisionTreeClassifier()
In [189]:
y_predDT = dt.predict(X_test)
In [190]:
accuracy_score(y_test,y_predDT)
Out[190]:
0.8019460138104206
In [209]:
precisionDT = precision_score(y_test, y_predDT)
print(f"The precision score is: {precisionDT:.3f}")

The precision score is: 0.785

Random Forest Classifier¶

In [192]:
from sklearn.ensemble import RandomForestClassifier
In [195]:
rf = RandomForestClassifier()
In [196]:
rf.fit(X_train,y_train)
Out[196]:
RandomForestClassifier()
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RandomForestClassifier()
In [197]:
y_predRF = rf.predict(X_test)
In [198]:
accuracy_score(y_test,y_predRF)
Out[198]:
0.869428750784683
In [210]:
precisionRF = precision_score(y_test, y_predRF)
print(f"The precision score is: {precisionRF:.3f}")

The precision score is: 0.868

Gradient Boosting Classifier¶

In [200]:
from sklearn.ensemble import GradientBoostingClassifier
In [201]:
gbc = GradientBoostingClassifier()
In [202]:
gbc.fit(X_train,y_train)
Out[202]:
GradientBoostingClassifier()
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GradientBoostingClassifier()
In [203]:
y_predGBC = gbc.predict(X_test)
In [204]:
accuracy_score(y_test,y_predGBC)
Out[204]:
0.8414940364092907
In [211]:
precisionGBC = precision_score(y_test, y_predGBC)
print(f"The precision score is: {precisionGBC:.3f}")

The precision score is: 0.837

-- Ending of model testing --

In [212]:
final_data = pd.DataFrame({'Models':['LR','SVC','KNN','DT','RF','GBC'],
                          'Accuracy':[accuracy_score(y_test,y_predict),
                                      accuracy_score(y_test,y_predSVC),
                                      accuracy_score(y_test,y_predKNN),
                                      accuracy_score(y_test,y_predDT),
                                      accuracy_score(y_test,y_predRF),
                                      accuracy_score(y_test,y_predGBC)],
                           'Precision Accuracy' : [precision,precisionSVC,precisionKNN,precisionDT,precisionRF,precisionGBC]})
In [213]:
final_data
Out[213]:
Models Accuracy Precision Accuracy
0 LR 0.779033 0.837194
1 SVC 0.846516 0.843226
2 KNN 0.822348 0.805952
3 DT 0.801946 0.784920
4 RF 0.869429 0.868438
5 GBC 0.841494 0.837194
In [216]:
# Create a bar graph for accuracy
colors = ['blue', 'green', 'red', 'purple']
plt.figure(figsize=(10, 5))
plt.bar(final_data['Models'], final_data['Accuracy'],color=colors, label='Accuracy')
plt.xlabel('Models')
plt.ylabel('Accuracy')
plt.title('Accuracy for Different Models')
plt.legend()
plt.show()
No description has been provided for this image
In [220]:
colors = ['blue', 'green', 'red', 'purple']
plt.figure(figsize=(10, 5))
plt.bar(final_data['Models'], final_data['Precision Accuracy'],color=colors, label='Precision')
plt.xlabel('Models')
plt.ylabel('Precision')
plt.title('Precision for Different Models')
plt.legend()
plt.show()

plt.figure(figsize=(6, 6))
plt.pie(final_data['Precision Accuracy'], labels=final_data['Models'], autopct='%1.1f%%', startangle=140)
plt.axis('equal')
plt.title('Precision for Different Models')
plt.show()
No description has been provided for this image
No description has been provided for this image

Final testing and Deployment¶

According to the accuracy reports the we will be going ahead with the Random Forest Model

In [221]:
X_res = sc.fit_transform(x_res)
In [222]:
rf.fit(X_res, y_res)
Out[222]:
RandomForestClassifier()
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RandomForestClassifier()
In [223]:
!pip3 install joblib --quiet

106273.17s - pydevd: Sending message related to process being replaced timed-out after 5 seconds

[notice] A new release of pip is available: 23.2.1 -> 23.3
[notice] To update, run: python3.11 -m pip install --upgrade pip
In [241]:
# Using joblib to save the model 
import joblib
joblib.dump(rf,'Bank_churn_model')
Out[241]:
['Bank_churn_model']

Implementing the model¶

In [242]:
modelcheck = joblib.load('Bank_churn_model')
In [243]:
df.columns
Out[243]:
Index(['CreditScore', 'Age', 'Tenure', 'Balance', 'NumOfProducts', 'HasCrCard',
       'IsActiveMember', 'EstimatedSalary', 'Exited', 'Geography_Germany',
       'Geography_Spain', 'Gender_Male'],
      dtype='object')

Outputs¶

In [258]:
result1 = modelcheck.predict([[619, 42, 2, 0.0, 0, 0, 0, 11.34, 0, 0, 0]])
if result == 1:
    print ("Customer won't churn")
else:
    print ("Customer will churn BEWARE")

Customer won't churn
In [263]:
result2 = modelcheck.predict([[789, 21, 1, 132424, 0, 1, 1, 1123423.34, 0, 0, 0]])
if result2 == 1:
    print ("Customer won't churn")
else:
    print ("Customer will churn BEWARE")

Customer will churn BEWARE
In [ ]: